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Let's say there is an algorithm that runs on a processor at x MIPS, operating at a clock frequency of y MHz. If I double the operating frequency to 2y, what happens to the MIPS? It seems to me that doubling the operating frequency should double the instructions executed per second, i.e, MIPS = 2x, is this correct?

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Only if you double the speed of everything. Which you can’t in practice.

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  • $\begingroup$ Since MIPS probably disregard memory or other I/O, I don't follow. $\endgroup$ – Raphael Dec 21 '17 at 9:48
  • $\begingroup$ MIPS don't have a choice. If you read from or write to memory, it takes as long as it takes. $\endgroup$ – gnasher729 Dec 21 '17 at 20:05
  • $\begingroup$ I was thinking of FLOPS which, afaik, are measured/calculated assuming no I/O happens. Which is purely theoretical, of course, but still. $\endgroup$ – Raphael Dec 21 '17 at 21:12
  • $\begingroup$ This answer would be fine if it had some justification. As always, we're looking for answers that explain why they're right. $\endgroup$ – David Richerby Dec 21 '17 at 22:11

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